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  • Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 25

Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 25

Answers (1)

Answer:

                k=6

Hint:

f\left ( x \right ) must be defined. The limit of the  f\left ( x \right ) approaches the value k must exist.

Given:

          f(x)=\left\{\begin{array}{c} \frac{k \cos x}{\pi-2 x}, \text { if } x \neq \frac{\pi}{2} \\\\ 3, \text { if } x=\frac{\pi}{2} \end{array}\right. 

Solution:

                f(x)=\left\{\begin{array}{c} \frac{k \cos x}{\pi-2 x}, \text { if } x \neq \frac{\pi}{2} \\\\ 3, \text { if } x=\frac{\pi}{2} \end{array}\right.

If f\left ( x \right ) is continuous at x=\frac{\pi}{2} , then

                \begin{aligned} &\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right) \\\\ &\lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}=3 \end{aligned}                                                                       ......(i)

                                                                                                                                

Putting x=\frac{\pi}{2}-h

                \lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}

From (i)    

                \begin{aligned} &\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}=3 \\\\ &\lim _{h \rightarrow 0} \frac{k \sinh }{2 h}=3 \\\\ &\lim _{h \rightarrow 0} \frac{k \sinh }{h}=6 \end{aligned}

                k \lim _{h \rightarrow 0} \frac{\sinh }{h}=6                                                                                     \left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]

                k\times 1= 6

                k=6, f\left ( x \right )  is continuous at x=\frac{\pi}{2} .

 

Posted by

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